"Correct" may not be the right question here. You are basically asking for
what Bayesian's call an "ignorance prior": a distribution that expresses
complete ignorance about some quantity. This is a difficult question.
Anyway...
One way to think of your problem is the following. Start with one interval.
The space of such intervals is a two-dimensional square, SO = [0,1] x [0,1],
in which lie the values [x, y] for the interval. Suppose that your
constraints specify a subset A of S. You would like a uniform distribution
on A. The probability of a point is then (underlying measure understood):
Pr([x, y] | A satisfied) = Pr([x, y] & A satisifed) / Pr(A satisfied) =
X_{A}(x, y) / |A|
where: |A| denotes the volume of A in the dxdy measure; X_{A} is the
characteristic function of A; and the volume of S has cancelled out (it is 1
anyway). If you want to sample, you simply sample x and y uniformly from
[0,1] (the measure on S is the product measure coming from [0,1]), and then
reject those that do not satisfy the constraints.
Now consider two intervals. The configuration space is now T = S^{2} =
[0,1]^{4}. You can sample points from this space by repeating the above
procedure, since the measure on T is just the product measure coming from S.
Note that the relevant portion of the space is given by A x B, where A and B
are the subsets of S coming from the two sets of constraints. So now you
have:
Pr([x,y], [w, z] | A & B satisfied) = Pr([x, y] & [w, z] & A & B satisfied)
/ Pr(A & B satisfied)
= X_{A x B}(x, y, w, z)
/ |A x B|
= X_{A}(x, y)X_{B}(w, z)
/ |A||B|
= Pr([x, y] | A
satisfied) Pr([w, z] | B satisfied)
as you would expect.
What about the intersection probability? If we denote the two intervals [x,
y] and [w, z], then the condition for intersection is:
intersection iff (x < z & w < y).
The intersection condition defines a subset V of T. The intersection
probability is then given by
Pr(intersection | constraints satisfied) = Pr(intersection & constraints
satisfied) / Pr(constraints satisfied)
= |V \cap (A x B)| /
|A x B|
where \cap denotes intersection and the volume of T has cancelled out (it is
1 anyway). So now there are two geometric problems: calculate the volumes of
V \cap (A x B) and of A x B. The latter is easy enough. It is just the
product |A||B|. Now A and B are trapezia in S, and their volumes turn out to
be:
|A| = (b - a)(2 - b - a)/2
where a and b are the limits on the length of the interval. This is for
ordered intervals, which are convenient because of the intersection
condition. Nothing would change if we considered unordered intervals except
that there would be extra factors of 2 everywhere that would eventually
cancel out.
The volume of the set V \cap (A x B) is considerably harder. I do not think
it is difficult in principle - it is a volume composed of trapezoids at
most - but it is very hard to visualize.
But maybe you had already come this far, in which case apologies for
redundancy.
Ian.
"Cristian-Augustin Saita" <[EMAIL PROTECTED]> wrote in
message asdbe7$qg7$[EMAIL PROTECTED]">news:asdbe7$qg7$[EMAIL PROTECTED]...
> Hi,
>
> In my experiments I choose (A,B) independent and uniform on (0,1), but
> I reject the cases where A > B (and respectively the cases where the
> size constraints are not satisfied). This seemed to me the most
> appropriate way to generate intervals uniformly distributed in (0,1),
> although I am not sure of its correctness.
>
> When no size constraints are imposed, the experimental evaluation of the
> intersection probability (generate a large number of interval pairs and
> verify how many of them overlap) by choosing (A,B) independent and
> uniform on (0,1) and rejecting the cases A > B seems to be equivalent to
> the third alternative below, namely choose (X,Y) independent and uniform
> on (0,1) and set A=min{X,Y} and B=max{X,Y}.
>
> I would appreciate if someone could tell me how I could generate points
> uniformly distributed in the set {(X,Y) | 0 <= X <= 1, 0 <= Y <= 1, X <=
> Y} and respectively in the set {(X,Y) | 0 <= X <= 1, 0 <= Y <= 1, X <=
> Y, a <= Y - X <= b, a, b constants, 0 <= a <= b <= 1}. Of course, all
> the four alternatives bellow, suggested by David, plus my proposition
> (based on rejection) could candidate for an answer, but which is the
> correct one?
>
> Thank you very much,
>
> Cristian Saita
>
> David Jones wrote:
> > "Cristian-Augustin Saita" <[EMAIL PROTECTED]> wrote in
> > message as86v0$9kh$[EMAIL PROTECTED]">news:as86v0$9kh$[EMAIL PROTECTED]...
> >
> >>I'm interested in the probability that two intervals selected "at
> >>random" from among those that satisfy the constraints will
> >
> > intersect.
> >
> >>"At random" here means that the intervals satisfying the constraints
> >>follow an uniform distribution.
> >>
> >>
> >
> >
> > Well maybe, but what do you think this should mean? The following four
> > ways of specifying an interval (A,B) yield different answers:
> >
> > (1) choose B uniformly on (0,1), then A uniformly on (0,A).
> >
> > (2) choose A uniformly on (0,1), then B uniformly on (A,1).
> >
> > (3) choose (X,Y) independent and uniform on (0,1), then set
> > A=min(X,Y), B=max(X,Y).
> >
> > (4) choose the length L to be uniform on (0,1), then choose A to be
> > uniform on (0,1-L), and set B=A+L.
> >
> > David Jones
> >
> >
>
.
.
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